allcoding1

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📈 Analytical overview of Telegram channel allcoding1

Channel allcoding1 (@allcoding1) in the English language segment is an active participant. Currently, the community unites 22 455 subscribers, ranking 8 839 in the Education category and 19 130 in the India region.

📊 Audience metrics and dynamics

Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 22 455 subscribers.

According to the latest data from 23 June, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -362 over the last 30 days and by -9 over the last 24 hours, overall reach remains high.

Verification status: Not verified
Engagement rate (ER): The average audience engagement rate is 5.09%. Within the first 24 hours after publication, content typically collects 1.29% reactions from the total number of subscribers.
Post reach: On average, each post receives 1 144 views. Within the first day, a publication typically gains 290 views.
Reactions and interaction: The audience actively supports content: the average number of reactions per post is 1.
Thematic interests: Content is focused on key topics such as dsa, stack, namaste, javascript, learning.

📝 Description and content policy

Channel description not provided.

Thanks to the high frequency of updates (latest data received on 24 June, 2026), the channel maintains relevance and a high level of publication reach. Analytics show that the audience actively interacts with content, making it an important point of influence in the Education category.

22 455

Subscribers

-924 hours

-737 days

-36230 days

1 144

Post views

~ 29024 hours

~ 30548 hours

5.09%

Engagement rate

No data

Posts per day

Ads index

beta

Posts Archive

22 455

magic Number Code python 3 All test cases passed Telegram:- @allcoding1

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XOR elements C++ language Telegram:- @allcoding1

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XOR code do you want

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C++ Question:- You are given Queries.each query consists of 3 integer X[i],Y[i] and N[i] where i donates the ith query The answer to ith query X[i],Y[i] and N[i] is denoted By (X[i],Y[i])^((X[i]+1),Y[i])^....^((X[i],N[i]*Y[i]) Find the sum of all answers modulo 10^9+7 Telegram:- @allcoding1

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You are given Queries.each query consists of 3 integer X[i],Y[i] and N[i] where i donates the ith query The answer to ith query X[i],Y[i] and N[i] is denoted By (X[i],Y[i])^((X[i]+1),Y[i])^....^((X[i],N[i]*Y[i]) Find the sum of all answers modulo 10^9+7

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#include <bits/stdc++.h> using namespace std; int count(string s) { int N, i, cnt = 0, ans = 0; N = s.length(); for (i = 0; i < N; i++) { if (s[i] == 'R') cnt++; if (s[i] == 'L') ans += cnt; } return ans; } int main() { string s = "RRLL"; cout << count(s) << endl; return 0; } C++ Telegram - @allcoding1

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def getLargestString(s, k): frequency_array = [0] * 26 for i in range(len(s)): frequency_array[ord(s[i]) - ord('a')] += 1 ans = "" i = 25 while i >= 0: if (frequency_array[i] > k): temp = k st = chr( i + ord('a')) while (temp > 0): ans += st temp -= 1 frequency_array[i] -= k j = i - 1 while (frequency_array[j] <= 0 and j >= 0): j -= 1 if (frequency_array[j] > 0 and j >= 0): str1 = chr(j + ord( 'a')) ans += str1 frequency_array[j] -= 1 else: break elif (frequency_array[i] > 0): temp = frequency_array[i] frequency_array[i] -= temp st = chr(i + ord('a')) while (temp > 0): ans += st temp -= 1 else: i -= 1 return ans if name == "main": S = input() k = 3 print (getLargestString(S, k)) Python Bob code Telegram:- @allcoding1

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Python Find Minimum Distance Between two numbers in Array code Telegram:- @allcoding1

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if (n > k && n > a) { printf("N is the Biggest number is %d", n); return n; } if (k > n && k > a) { printf("k is the Biggest number is %d", k); return k; } if (a > n && a > k) { printf("a is the Biggest number is %d", a); return a; } Telegram:- @allcoding1

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from itertools import permutations def count(arr): z=[] perm = permutations(arr) for i in list(perm): z.append(list(i)) q=[] for i in range(len(arr)-1): x,y=arr[i],arr[i+1] for j in range(len(z)): if z[j].index(x)!=len(z[j])-1: if z[j][z[j].index(x)+1]==y: q.append(z[j]) for i in range(len(q)): if q[i] in z: z.remove(q[i]) return len(z) a= int(input()) b=list(map(int,input().strip().split())) print(count(b)) Python Telegram - @allcoding1

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Java language Telegram:- @allcoding1

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Java language Telegram:- @allcoding1

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Python 3 Telegram:- @allcoding1

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Infosys exam Answer's (22/10/22) https://www.joboffersadda.com/2022/01/infosys-exam-answer.html?m=1 https://www.joboffersadda.com/2022/01/infosys-exam-answer.html?m=1 https://www.joboffersadda.com/2022/01/infosys-exam-answer.html?m=1 Telegram:- @Infosys_Ans

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Python 3 Once check Output Telegram:- @allcoding1

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infosys exam 22oct_discussion_group @allcoding1 @allcoding1 @allcoding1 @allcoding1 @allcoding1 @allcoding1